2009
2009
2009
2009 Utah
Utah
Utah
Utah State
State
State Mathematics Contest
State
Mathematics Contest
Mathematics Contest
Mathematics Contest
Senior Exam
Senior Exam
Senior Exam
Senior Exam Solutions
Solutions
Solutions
Solutions
1. (e) 56.25 1. (e) 56.25 1. (e) 56.25 1. (e) 56.25
V = V = V =
V = π rπ rπ rπ r2222 h. If the diameter is down 20%, the radius is down 20%. To maintain the same volume, the equation h. If the diameter is down 20%, the radius is down 20%. To maintain the same volume, the equation h. If the diameter is down 20%, the radius is down 20%. To maintain the same volume, the equation h. If the diameter is down 20%, the radius is down 20%. To maintain the same volume, the equation
describing the adjusted lengths would be describing the adjusted lengths would be describing the adjusted lengths would be
describing the adjusted lengths would be V = V = V = V = π( π( π( π( -. r)r)r)r)2222 ( ( ( ( /.
01 h). Since h). Since h). Since h). Since /.
01 = 1.5625, there must be an increase of = 1.5625, there must be an increase of = 1.5625, there must be an increase of = 1.5625, there must be an increase of
56.25%. 56.25%. 56.25%. 56.25%. 2.
2. 2. 2. (d) 3(d) 3(d) 3(d) 3
Let X be the initial number of coins. Since their average value is 20 cents, their total value is 20X. By adding a Let X be the initial number of coins. Since their average value is 20 cents, their total value is 20X. By adding a Let X be the initial number of coins. Since their average value is 20 cents, their total value is 20X. By adding a Let X be the initial number of coins. Since their average value is 20 cents, their total value is 20X. By adding a quarter, we get
quarter, we get quarter, we get
quarter, we get /78 9 /.8 9 0 = 21. Solving for X, th= 21. Solving for X, th= 21. Solving for X, th= 21. Solving for X, there were originally 4 coins, totaling 80 cents in value. They must ere were originally 4 coins, totaling 80 cents in value. They must ere were originally 4 coins, totaling 80 cents in value. They must ere were originally 4 coins, totaling 80 cents in value. They must have been three quarters and a nickel.
have been three quarters and a nickel. have been three quarters and a nickel. have been three quarters and a nickel. 3. (e) p
3. (e) p 3. (e) p 3. (e) p2222 > > > > ----4q4q4q4q
The discriminant of the quadratic formula is (bThe discriminant of the quadratic formula is (bThe discriminant of the quadratic formula is (bThe discriminant of the quadratic formula is (b2222 –––– 4ac), which in this case is (p4ac), which in this case is (p4ac), which in this case is (p4ac), which in this case is (p2222 + 4q). + 4q). + 4q). + 4q). ForForFor a quadratic equation to Fora quadratic equation to a quadratic equation to a quadratic equation to
have have have
have two two two distinct real roots, ptwo distinct real roots, pdistinct real roots, pdistinct real roots, p2222 + 4q > 0, or p+ 4q > 0, or p+ 4q > 0, or p+ 4q > 0, or p2222 > > > ----4q.> 4q.4q.4q.
4444. . . . (c) 8(c) 8(c) 8(c) 8
CDE/F ∙∙∙∙ CDEH/. ∙∙∙∙ CDE.- = = = = CDE /CDE F ∙∙∙∙ CDE /.CDE H ∙∙∙∙ CDE -CDE . = = = = / CDE HCDE / ∙∙∙∙ / CDE /.CDE H ∙∙∙∙ / CDE /CDE . = 2= 2= 2= 2 ∙∙∙∙ 2 2 2 2 ∙ 2 = 8∙ 2 = 8∙ 2 = 8∙ 2 = 8
5555. . . . (a) 6(a) 6(a) 6(a) 6
For aFor aFor aFor annnn integer to be integer to be integer to be integer to be both a both a both a both a square and a cube, it must be square and a cube, it must be a perfectsquare and a cube, it must be square and a cube, it must be a perfecta perfect sixth power. a perfectsixth power. sixth power. Positive intsixth power. Positive intPositive integers to the sixth power Positive integers to the sixth power egers to the sixth power egers to the sixth power include
include include
include 01 = 1; = 1; = 1; = 1; /1 = 64; = 64; H= 64; = 64; 1 = 729; = 729; = 729; -= 729; 1 = 4096; = 4096; .= 4096; = 4096; 1 = 15625; = 15625; = 15625; 1= 15625; 1 = 46656; = 46656; L= 46656; = 46656; 1 = 117649… only the first = 117649… only the first = 117649… only the first six= 117649… only the first sixsixsix
positive positive positive
positive integers integers integers integers to the sixth power are less than 100,000.to the sixth power are less than 100,000.to the sixth power are less than 100,000. to the sixth power are less than 100,000. 6.
6. 6. 6. (b) 240(b) 240(b) 240(b) 240
Let X be the number of voters from Salt Lake County, and Y be the number of voters from Utah County, both Let X be the number of voters from Salt Lake County, and Y be the number of voters from Utah County, both Let X be the number of voters from Salt Lake County, and Y be the number of voters from Utah County, both Let X be the number of voters from Salt Lake County, and Y be the number of voters from Utah County, both measure
measure measure
measuredddd in thousands. X + Y = 840, and 0.65X + 0.72Y = 0.67(840). Solving this system of equations yields that in thousands. X + Y = 840, and 0.65X + 0.72Y = 0.67(840). Solving this system of equations yields that in thousands. X + Y = 840, and 0.65X + 0.72Y = 0.67(840). Solving this system of equations yields that in thousands. X + Y = 840, and 0.65X + 0.72Y = 0.67(840). Solving this system of equations yields that X = 600, Y = 240.
X = 600, Y = 240. X = 600, Y = 240. X = 600, Y = 240. 7777. . . . (e) 36(e) 36(e) 36(e) 36
The minimum number of sections is 10 (draw The minimum number of sections is 10 (draw The minimum number of sections is 10 (draw
The minimum number of sections is 10 (draw 9999 parallel lines). To achieve maximum number of sections, each parallel lines). To achieve maximum number of sections, each parallel lines). To achieve maximum number of sections, each parallel lines). To achieve maximum number of sections, each successive line must cross all existing lines. The first line creates 2 sections; the second line cross the first and successive line must cross all existing lines. The first line creates 2 sections; the second line cross the first and successive line must cross all existing lines. The first line creates 2 sections; the second line cross the first and successive line must cross all existing lines. The first line creates 2 sections; the second line cross the first and leads to 4 sections; the third line crosses the first 2 and leads 7 sections
leads to 4 sections; the third line crosses the first 2 and leads 7 sections leads to 4 sections; the third line crosses the first 2 and leads 7 sections
leads to 4 sections; the third line crosses the first 2 and leads 7 sections. The N. The N. The N. The Nthththth new line can create N new new line can create N new new line can create N new new line can create N new
sections. Beginning with 1 section (the plane), and adding 1 + 2 + … + 8 + 9 new sections gives 46 maximum sections. Beginning with 1 section (the plane), and adding 1 + 2 + … + 8 + 9 new sections gives 46 maximum sections. Beginning with 1 section (the plane), and adding 1 + 2 + … + 8 + 9 new sections gives 46 maximum sections. Beginning with 1 section (the plane), and adding 1 + 2 + … + 8 + 9 new sections gives 46 maximum sections. 46
sections. 46 sections. 46
sections. 46 –––– 10 = 36.10 = 36.10 = 36. 10 = 36. 8888. . . . (c) 116(c) 116(c) 116(c) 116
The center of mass of the left weight is 2.5 steps away from the fulcrum, The center of mass of the left weight is 2.5 steps away from the fulcrum, The center of mass of the left weight is 2.5 steps away from the fulcrum,
The center of mass of the left weight is 2.5 steps away from the fulcrum, while the center of mass of the right while the center of mass of the right while the center of mass of the right while the center of mass of the right weight is 14.5 steps away. It must be the case that X
weight is 14.5 steps away. It must be the case that X weight is 14.5 steps away. It must be the case that X
weight is 14.5 steps away. It must be the case that X ∙ 2.5 = 20 ∙ 14.5. Solving for X gives 116.∙ 2.5 = 20 ∙ 14.5. Solving for X gives 116.∙ 2.5 = 20 ∙ 14.5. Solving for X gives 116. ∙ 2.5 = 20 ∙ 14.5. Solving for X gives 116. 9999. . . . (d) 5(d) 5(d) 5(d) 5
There are three possible situations: (I) the exponent polynomial is equal to zero, (II) the base polynomial iThere are three possible situations: (I) the exponent polynomial is equal to zero, (II) the base polynomial iThere are three possible situations: (I) the exponent polynomial is equal to zero, (II) the base polynomial iThere are three possible situations: (I) the exponent polynomial is equal to zero, (II) the base polynomial is equal s equal s equal s equal to one, or (III) the base polynomial is equal to (
to one, or (III) the base polynomial is equal to ( to one, or (III) the base polynomial is equal to (
to one, or (III) the base polynomial is equal to (----1) and the exponent is an even integer. For (I), the two solutions 1) and the exponent is an even integer. For (I), the two solutions 1) and the exponent is an even integer. For (I), the two solutions 1) and the exponent is an even integer. For (I), the two solutions are
are are
are ((((----1) and 1) and 1) and 1) and RS0-T. For (II), the two solutions are 3 and . For (II), the two solutions are 3 and . For (II), the two solutions are 3 and . For (II), the two solutions are 3 and -H. For (III), both 1 and . For (III), both 1 and . For (III), both 1 and . For (III), both 1 and 00- would set the base exponential to would set the base exponential to would set the base exponential to would set the base exponential to ((((----1111), but ), but ), but ), but 00- would make the exponent an odd integer. So, the 5 solutions are: would make the exponent an odd integer. So, the 5 solutions are: would make the exponent an odd integer. So, the 5 solutions are: would make the exponent an odd integer. So, the 5 solutions are: US0, S0-, H-, 0, HV....
10. (a) 10. (a) 10. (a) 10. (a) 1WH X HYLY X 1WH
A A A A regular hexagon can be dividedregular hexagon can be dividedregular hexagon can be divided into six equilateral triangles.regular hexagon can be dividedinto six equilateral triangles.into six equilateral triangles.into six equilateral triangles. Each side length is equal to the larger Each side length is equal to the larger Each side length is equal to the larger Each side length is equal to the larger radius, R. The height ofradius, R. The height ofradius, R. The height ofradius, R. The height of
each equilateral triangle is the smaller radius, equal to each equilateral triangle is the smaller radius, equal to each equilateral triangle is the smaller radius, equal to each equilateral triangle is the smaller radius, equal to WH/R. The shadedR. The shaded R. The shadedR. The shaded area is equal to the difference between the area of the hexagonarea is equal to the difference between the area of the hexagonarea is equal to the difference between the area of the hexagonarea is equal to the difference between the area of the hexagon
and the area of the smaller circle = 6 and the area of the smaller circle = 6 and the area of the smaller circle = 6
and the area of the smaller circle = 6 ∙ ∙ ∙ ∙ 0/ ∙∙∙∙ WH/ RRRR2222 –––– H\
-RRRR2222. The unshaded area. The unshaded area. The unshaded area. The unshaded area
is equal to the area of the large circle minus the shaded area =is equal to the area of the large circle minus the shaded area =is equal to the area of the large circle minus the shaded area =is equal to the area of the large circle minus the shaded area = πRπRπRπR2222 –––– (6 (6 (6 (6 ∙ ∙ ∙ ∙ 0
/ ∙∙∙∙ WH/ RRRR2222 –––– H\
-RRRR2222). The ratio of the two areas is, when simplified,). The ratio of the two areas is, when simplified, ). The ratio of the two areas is, when simplified,). The ratio of the two areas is, when simplified,
equal to equal to equal to equal to 1WH X HYLY X 1WH
11. 11. 11.
11. (a) (a) (a) ]01 H/(a) 7 01^
_-is equal tois equal tois equal tois equal to `_/a/. By matrix multiplication, _. By matrix multiplication, . By matrix multiplication, . By matrix multiplication, / = = = = ]
-7 -^. This result squared is . This result squared is . This result squared is . This result squared is ]01 H/7 01^.... 12
12 12
12. . . . (d) 126(d) 126(d) 126 (d) 126
In the binomial expansion of In the binomial expansion of In the binomial expansion of
In the binomial expansion of (b + c)F, what is the coefficient of the x, what is the coefficient of the x, what is the coefficient of the x, what is the coefficient of the x4444yyyy5555 term?term?term?term?
The NThe NThe NThe Nthththth term of an Mterm of an Mterm of an Mterm of an Mthththth----degree binomial expansion is degree binomial expansion is degree binomial expansion is degree binomial expansion is R e
11113333. . . . (d) 500(d) 500(d) 500 (d) 500
The number of zero The number of zero The number of zero
The number of zero----digits at the end of (2009g) will be decided by the number of 2’s and 5’s which are factors of digits at the end of (2009g) will be decided by the number of 2’s and 5’s which are factors of digits at the end of (2009g) will be decided by the number of 2’s and 5’s which are factors of digits at the end of (2009g) will be decided by the number of 2’s and 5’s which are factors of (2009g). With quick consideration, there will be fewer 5’s than 2’s.
(2009g). With quick consideration, there will be fewer 5’s than 2’s. (2009g). With quick consideration, there will be fewer 5’s than 2’s.
(2009g). With quick consideration, there will be fewer 5’s than 2’s. Less than 2009, there are 401Less than 2009, there are 401Less than 2009, there are 401 multiples of 5, 80 Less than 2009, there are 401multiples of 5, 80 multiples of 5, 80 multiples of 5, 80 multiples of 25, 16 multiples of 125, and 3 multiples of 6
multiples of 25, 16 multiples of 125, and 3 multiples of 6 multiples of 25, 16 multiples of 125, and 3 multiples of 6
multiples of 25, 16 multiples of 125, and 3 multiples of 625. Thus, there are will be (40125. Thus, there are will be (40125. Thus, there are will be (401 + 80 + 16 + 3) = 500 25. Thus, there are will be (401+ 80 + 16 + 3) = 500 + 80 + 16 + 3) = 500 + 80 + 16 + 3) = 500 factors of 5, and at least 500 factors of 2 which divide (2009g).
factors of 5, and at least 500 factors of 2 which divide (2009g). factors of 5, and at least 500 factors of 2 which divide (2009g). factors of 5, and at least 500 factors of 2 which divide (2009g). 14. (c) 640
14. (c) 640 14. (c) 640 14. (c) 640
TideTideTideTide speed = X. Speed with tspeed = X. Speed with tspeed = X. Speed with tidespeed = X. Speed with tideideide = (100 + X). Speed against = (100 + X). Speed against = (100 + X). Speed against tide= (100 + X). Speed against tide = (100 tidetide= (100 = (100 = (100 –––– X).X).X).X). 4(100 + X) = 16(100 4(100 + X) = 16(100 4(100 + X) = 16(100 4(100 + X) = 16(100 –––– X). X). X). X). Solving for X gives that X = 60. Swimming 4 minutes at 160 yards per minute will move Greg 640 yards.
Solving for X gives that X = 60. Swimming 4 minutes at 160 yards per minute will move Greg 640 yards. Solving for X gives that X = 60. Swimming 4 minutes at 160 yards per minute will move Greg 640 yards. Solving for X gives that X = 60. Swimming 4 minutes at 160 yards per minute will move Greg 640 yards. 15
15 15
15. . . . (c) (c) (c) –––– 2c(c) 2c2c2c Since Since Since
Since lml b = = (mno b)= = X0 , the , the , the expression , the expression expression CDEexpression p(lml/b) is equal to (is equal to (----2)is equal to (is equal to (2)2) CDE2) p(mno b) = (= (= (= (----2) 2) 2) 2) ∙ c.∙ c.∙ c. ∙ c.
16 16 16
16. . . . (b) (b) (b) (b) W. 9 0/
Examine the complex fraction as a matter of repetition. B = 1 + Examine the complex fraction as a matter of repetition. B = 1 + Examine the complex fraction as a matter of repetition. B = 1 +
Examine the complex fraction as a matter of repetition. B = 1 + 0 9 qq ; by clearing denominators, you get B(1 + B) ; by clearing denominators, you get B(1 + B) ; by clearing denominators, you get B(1 + B) ; by clearing denominators, you get B(1 + B) = (1 + B) + B. Put into standard
= (1 + B) + B. Put into standard = (1 + B) + B. Put into standard
= (1 + B) + B. Put into standard form, this is the quadratic equation Bform, this is the quadratic equation Bform, this is the quadratic equation Bform, this is the quadratic equation B2222 –––– B + 1 = 0. Solving for B gives B + 1 = 0. Solving for B gives B + 1 = 0. Solving for B gives B + 1 = 0. Solving for B gives 0 r W. / . . . .
However, since B is clearly greater than one, the answer must be However, since B is clearly greater than one, the answer must be However, since B is clearly greater than one, the answer must be However, since B is clearly greater than one, the answer must be 0 9 W. / . . . . 11117777. . . . (b) 5(b) 5(b) 5 (b) 5
The Remainder Theorem states The Remainder Theorem states The Remainder Theorem states
The Remainder Theorem states that if a function, f(x) is divided by (x that if a function, f(x) is divided by (x that if a function, f(x) is divided by (x that if a function, f(x) is divided by (x –––– c), the remainder is f(c). Evaluating the c), the remainder is f(c). Evaluating the c), the remainder is f(c). Evaluating the c), the remainder is f(c). Evaluating the expression at x =
expression at x = expression at x =
expression at x = ----1, we get 1, we get 1, we get 1, we get 13 13 13 –––– ((((----6666)))) + 13 + + + ((((----12121212)))) –––– ((((----3333)))) –––– 5555 = 5.= 5.= 5.= 5. 11118888.... (e) (e) (e) ----10(e) 101010
If the determinant of A is equal to 0, then If the determinant of A is equal to 0, then If the determinant of A is equal to 0, then If the determinant of A is equal to 0, then _X0 does not exist. The determinant of A is (xdoes not exist. The determinant of A is (xdoes not exist. The determinant of A is (xdoes not exist. The determinant of A is (x2222 + 10x + 10x + 10x –––– 24), which is + 10x 24), which is 24), which is 24), which is
equal to zero when x = equal to zero when x = equal to zero when x =
equal to zero when x = ----12 or 2. The sum of these results is 12 or 2. The sum of these results is 12 or 2. The sum of these results is ----10.12 or 2. The sum of these results is 10.10.10. 19.
19. 19. 19. (a) 8(a) 8(a) 8 (a) 8
Using common bases, Using common bases, Using common bases,
Using common bases, Lt = = = = L/u90/ and and and and /.u = = /= = -tXH1. So, y = 2x + 12, and 5x = 4y . So, y = 2x + 12, and 5x = 4y . So, y = 2x + 12, and 5x = 4y . So, y = 2x + 12, and 5x = 4y –––– 36. Solv36. Solv36. Solv36. Solving by substitution, ing by substitution, ing by substitution, ing by substitution,
x = x = x =
x = ----4 and y = 4. The difference between x and y is 8.4 and y = 4. The difference between x and y is 8.4 and y = 4. The difference between x and y is 8.4 and y = 4. The difference between x and y is 8.
20 20 20
20. . . . (a) (a) (a) (a) WH/ –––– 0/ iiii
For complex number z = a + bi, the equivalent value in the complex plane is |z| For complex number z = a + bi, the equivalent value in the complex plane is |z| For complex number z = a + bi, the equivalent value in the complex plane is |z|
For complex number z = a + bi, the equivalent value in the complex plane is |z| ∙ (cos ∙ (cos ∙ (cos ∙ (cos ϴϴϴϴ + i sin + i sin + i sin + i sin ϴϴϴϴ) = ) = ) = ) = |z|
|z| |z|
|z| ∙ ∙ ∙ ∙ xny. . . . As |z| = 1 for all choices, and as As |z| = 1 for all choices, and as As |z| = 1 for all choices, and as weAs |z| = 1 for all choices, and as wewewe need zneed zneed zneed z6666 = = ----1, then = = 1, then 1, then `x1, then nya1 = = ----1. The corresponding = = 1. The corresponding 1. The corresponding 1. The corresponding ϴϴϴϴ’s to the five ’s to the five ’s to the five ’s to the five
choices choices choices
choices are are are are X\1, , , , \H, , , , X\-, , , , \-, and , and , and , and X\H. The only . The only . The only . The only ϴϴϴϴ which will satisfy which will satisfy xwhich will satisfy which will satisfy 1ny = = = = ----1111 is is is is X\ 1 ....
22221111. . . . (d) 3731(d) 3731(d) 3731 (d) 3731 3G 3G 3G
3G2222 + 5G + 9 = 2009, or 3G+ 5G + 9 = 2009, or 3G+ 5G + 9 = 2009, or 3G+ 5G + 9 = 2009, or 3G2222 = = = ----5G + 2000. Since 5 divides= 5G + 2000. Since 5 divides ----5G + 2000, it must divide 3G5G + 2000. Since 5 divides5G + 2000. Since 5 divides 5G + 2000, it must divide 3G5G + 2000, it must divide 3G5G + 2000, it must divide 3G2222. So, G must be a multiple . So, G must be a multiple . So, G must be a multiple . So, G must be a multiple
of 5. Checking small multiples of 5 for 3G of 5. Checking small multiples of 5 for 3G of 5. Checking small multiples of 5 for 3G
of 5. Checking small multiples of 5 for 3G2222 + 5G = 2000, + 5G = 2000, + 5G = 2000, + 5G = 2000, G = 25 is a quick solution. So, it must be the case that M = G = 25 is a quick solution. So, it must be the case that M = G = 25 is a quick solution. So, it must be the case that M = G = 25 is a quick solution. So, it must be the case that M =
8. Since 8 8. Since 8 8. Since 8
8. Since 80000 = 1, 8= 1, 8= 1, 8= 1, 81111 = 8, 8= 8, 8= 8, 8= 8, 82222 = 64, 8= 64, 8= 64, 8= 64, 83333 = 512, it is easy to see that 8= 512, it is easy to see that 8= 512, it is easy to see that 8= 512, it is easy to see that 83333 goes into goes into goes into 2009 no more than 3 times, so the goes into 2009 no more than 3 times, so the 2009 no more than 3 times, so the 2009 no more than 3 times, so the
answer must be a four answer must be a four answer must be a four
answer must be a four----digit number starting with 3. Checking, 2009 is, indeed, equal to 3 digit number starting with 3. Checking, 2009 is, indeed, equal to 3 digit number starting with 3. Checking, 2009 is, indeed, equal to 3 ∙ 512 + 7 ∙ 64 + 3 ∙ 8 + 1.digit number starting with 3. Checking, 2009 is, indeed, equal to 3 ∙ 512 + 7 ∙ 64 + 3 ∙ 8 + 1.∙ 512 + 7 ∙ 64 + 3 ∙ 8 + 1. ∙ 512 + 7 ∙ 64 + 3 ∙ 8 + 1. 22222222. . . . (e) (e) (e) (e) /0-
Given that each brother rolls either an odd or an even number, there would be 2Given that each brother rolls either an odd or an even number, there would be 2Given that each brother rolls either an odd or an even number, there would be 2Given that each brother rolls either an odd or an even number, there would be 26666 = 64 outcomes from the result of = 64 outcomes from the result of = 64 outcomes from the result of = 64 outcomes from the result of
their rolls. However, since it is known that there are at their rolls. However, since it is known that there are at their rolls. However, since it is known that there are at
their rolls. However, since it is known that there are at least 3 even numbers, then the number of ways that exactly least 3 even numbers, then the number of ways that exactly least 3 even numbers, then the number of ways that exactly least 3 even numbers, then the number of ways that exactly 3, 4, 5, or 6 evens can be rolled is found by adding
3, 4, 5, or 6 evens can be rolled is found by adding 3, 4, 5, or 6 evens can be rolled is found by adding
3, 4, 5, or 6 evens can be rolled is found by adding R1HT + + + + R1-T + + + R1.T + + + + R+ 11T = 20 + 15 + 6 + 1 = 42. Gi= 20 + 15 + 6 + 1 = 42. Gi= 20 + 15 + 6 + 1 = 42. Given that = 20 + 15 + 6 + 1 = 42. Given that ven that ven that exactly 3, 4, 5, or 6 evens are rolled, the number of ways that Largo, Milo, and Paulo can possess three of those exactly 3, 4, 5, or 6 evens are rolled, the number of ways that Largo, Milo, and Paulo can possess three of those exactly 3, 4, 5, or 6 evens are rolled, the number of ways that Largo, Milo, and Paulo can possess three of those exactly 3, 4, 5, or 6 evens are rolled, the number of ways that Largo, Milo, and Paulo can possess three of those evens is found by calculating whom, of Kojo, Nico, and Otto, does not have an even number, adding
evens is found by calculating whom, of Kojo, Nico, and Otto, does not have an even number, adding evens is found by calculating whom, of Kojo, Nico, and Otto, does not have an even number, adding
evens is found by calculating whom, of Kojo, Nico, and Otto, does not have an even number, adding RHHT + + + + RH/T + + + + RH0T + + + + RH7T = 1 + 3 += 1 + 3 += 1 + 3 += 1 + 3 + 3 + 1 = 8. Thus, in 8 of the possible 42 scenarios in which there are at least 3 even rolls, 3 + 1 = 8. Thus, in 8 of the possible 42 scenarios in which there are at least 3 even rolls, 3 + 1 = 8. Thus, in 8 of the possible 42 scenarios in which there are at least 3 even rolls, 3 + 1 = 8. Thus, in 8 of the possible 42 scenarios in which there are at least 3 even rolls, Largo, Milo, and Paulo all have even rolls.
Largo, Milo, and Paulo all have even rolls. Largo, Milo, and Paulo all have even rolls. Largo, Milo, and Paulo all have even rolls.
22223333. . . . (b) 30 + 2(b) 30 + 2(b) 30 + 2W.07 (b) 30 + 2
By the Pythagorean Theorem, the shown diagonal must have By the Pythagorean Theorem, the shown diagonal must have By the Pythagorean Theorem, the shown diagonal must have By the Pythagorean Theorem, the shown diagonal must have length equal to 13.length equal to 13.length equal to 13.length equal to 13. Since the height and base of the right
Since the height and base of the right Since the height and base of the right
Since the height and base of the right triangle are known (with a little rotation),triangle are known (with a little rotation),triangle are known (with a little rotation),triangle are known (with a little rotation), the area of the
the area of the the area of the
the area of the right triangle is 30 squareright triangle is 30 squareright triangle is 30 squareright triangle is 30 square units. The area of the otherunits. The area of the otherunits. The area of the otherunits. The area of the other triangle
triangle triangle
triangle is calculated by Heron’sis calculated by Heron’sis calculated by Heron’s formula. Let p be the perimeter of ais calculated by Heron’sformula. Let p be the perimeter of aformula. Let p be the perimeter of aformula. Let p be the perimeter of a triangle. Let s be half of
triangle. Let s be half of triangle. Let s be half of
triangle. Let s be half of pppp (call it the semi(call it the semi(call it the semi(call it the semi----perimeter or the triangle).perimeter or the triangle).perimeter or the triangle). perimeter or the triangle). Heron’s formula states
Heron’s formula states Heron’s formula states
Heron’s formula states that, for any trianglthat, for any trianglthat, for any triangle with sides a, b, c, tthat, for any triangle with sides a, b, c, te with sides a, b, c, te with sides a, b, c, the areahe areahe areahe area is
isis
is calculated by the following formula:calculated by the following formula:calculated by the following formula: calculated by the following formula:
A = A = A = A = |m(m S p)(m S })(m S l) = = = W0L ∙ H ∙ - ∙ 07 = 2= = 2= 2= 2W.07....
24 24 24
24. . . . (b) 193(b) 193(b) 193 (b) 193
Let X be the amount of time after 3 PM when the minute hand crosses the Let X be the amount of time after 3 PM when the minute hand crosses the Let X be the amount of time after 3 PM when the minute hand crosses the Let X be the amount of time after 3 PM when the minute hand crosses the hourhourhour hand, measured in minutes. The hourhand, measured in minutes. The hand, measured in minutes. The hand, measured in minutes. The equation to find X is: X = 15 +
equation to find X is: X = 15 + equation to find X is: X = 15 +
equation to find X is: X = 15 + 178 ∙∙∙∙ 5. Solving, X = 5. Solving, X = 5. Solving, X = 5. Solving, X = 0h700, or 16 , or 16 , or 16 , or 16 00-, or about 16 minutes, 22 seconds. The second hand , or about 16 minutes, 22 seconds. The second hand , or about 16 minutes, 22 seconds. The second hand , or about 16 minutes, 22 seconds. The second hand overlaps the m
overlaps the m overlaps the m
overlaps the minute hand 59 times per hour. From noon to 3 o’clock, that makes 177 overlaps. From 3PM to 3:16 inute hand 59 times per hour. From noon to 3 o’clock, that makes 177 overlaps. From 3PM to 3:16 inute hand 59 times per hour. From noon to 3 o’clock, that makes 177 overlaps. From 3PM to 3:16 inute hand 59 times per hour. From noon to 3 o’clock, that makes 177 overlaps. From 3PM to 3:16 PM, the second hand will overlap the minute hand 15 more times. At 3:16:22, the second hand will have passed PM, the second hand will overlap the minute hand 15 more times. At 3:16:22, the second hand will have passed PM, the second hand will overlap the minute hand 15 more times. At 3:16:22, the second hand will have passed PM, the second hand will overlap the minute hand 15 more times. At 3:16:22, the second hand will have passed over the minute hand one additional time (at about
over the minute hand one additional time (at about over the minute hand one additional time (at about
over the minute hand one additional time (at about 3:16:16).3:16:16).3:16:16).3:16:16). 25.
25. 25.
25. (e) (e) (e) (e) 0777 lDm -hlDm 11
Because of the nature of supplementary angles, the obtuse angle opposite M measures 138°. By the Law of Sines, Because of the nature of supplementary angles, the obtuse angle opposite M measures 138°. By the Law of Sines, Because of the nature of supplementary angles, the obtuse angle opposite M measures 138°. By the Law of Sines, Because of the nature of supplementary angles, the obtuse angle opposite M measures 138°. By the Law of Sines,
mno 0Hh
• = = = = mno /-0777. Solving for M, we get that M = . Solving for M, we get that M = . Solving for M, we get that M = . Solving for M, we get that M = 0777 mno 0Hhmno /- . Now, trigonometric identities verify that sin . Now, trigonometric identities verify that sin ϴϴϴϴ = sin (180 . Now, trigonometric identities verify that sin . Now, trigonometric identities verify that sin = sin (180 = sin (180 = sin (180
–––– ϴϴϴϴ) , and that sin ) , and that sin ) , and that sin ϴϴϴϴ = cos (90 ) , and that sin = cos (90 = cos (90 = cos (90 ---- ϴϴϴϴ), where 0 ), where 0 ), where 0 ), where 0 ≤ ≤ ≤ ≤ ϴϴϴϴ ≤ 90. Using these properties, sin 138 = sin 42 = cos 48, and sin 24 ≤ 90. Using these properties, sin 138 = sin 42 = cos 48, and sin 24 ≤ 90. Using these properties, sin 138 = sin 42 = cos 48, and sin 24 ≤ 90. Using these properties, sin 138 = sin 42 = cos 48, and sin 24 = cos 66. By replacement, we get
= cos 66. By replacement, we get = cos 66. By replacement, we get
= cos 66. By replacement, we get 0777 lDm -hlDm 11 .... 22226666. . . . (c) (c) (c) 410(c) 410410410
Let A be the smallest member of the sequence, and let D be the difference between consecutive terms. In an Let A be the smallest member of the sequence, and let D be the difference between consecutive terms. In an Let A be the smallest member of the sequence, and let D be the difference between consecutive terms. In an Let A be the smallest member of the sequence, and let D be the difference between consecutive terms. In an arithmetic sequence, the mean and median are the same number, so the 4
arithmetic sequence, the mean and median are the same number, so the 4 arithmetic sequence, the mean and median are the same number, so the 4
arithmetic sequence, the mean and median are the same number, so the 4thththth term is equal to one seventh of 2009, term is equal to one seventh of 2009, term is equal to one seventh of 2009, term is equal to one seventh of 2009,
or 287. or 287. or 287.
or 287. We have that A + 3D = 287We have that A + 3D = 287We have that A + 3D = 287We have that A + 3D = 287, and A + (A + D) = , and A + (A + D) = , and A + (A + D) = , and A + (A + D) = 0H (A + 4D + A + 5D + A + 6D). Solving the system of (A + 4D + A + 5D + A + 6D). Solving the system of (A + 4D + A + 5D + A + 6D). Solving the system of (A + 4D + A + 5D + A + 6D). Solving the system of equations, we get that A = 164 and D = 41. Thus, the largest element of the sequence, (A + 6D) = 164 + 6 equations, we get that A = 164 and D = 41. Thus, the largest element of the sequence, (A + 6D) = 164 + 6 equations, we get that A = 164 and D = 41. Thus, the largest element of the sequence, (A + 6D) = 164 + 6 equations, we get that A = 164 and D = 41. Thus, the largest element of the sequence, (A + 6D) = 164 + 6 ∙ 41 = ∙ 41 = ∙ 41 = ∙ 41 = 410.
410. 410. 410. 22227777. . . . (a) 0 to 0(a) 0 to 0(a) 0 to 0 (a) 0 to 0
From the tournament points, NE must have had From the tournament points, NE must have had From the tournament points, NE must have had
From the tournament points, NE must have had 2222 wins and wins and wins and wins and 1111 tie. LA must have had tie. LA must have had tie. LA must have had tie. LA must have had 2222 losses and losses and losses and losses and 1111 tie. tie. tie. tie. NE and LA NE and LA NE and LA NE and LA could not have tied each other, because NE must have scored at least
could not have tied each other, because NE must have scored at least could not have tied each other, because NE must have scored at least
could not have tied each other, because NE must have scored at least 2222 goals against LA (since at most 3 goals were goals against LA (since at most 3 goals were goals against LA (since at most 3 goals were goals against LA (since at most 3 goals were scored against LA by DC and RSL).
scored against LA by DC and RSL). scored against LA by DC and RSL).
scored against LA by DC and RSL). Since NE and LA did not tie each other, thSince NE and LA did not tie each other, thSince NE and LA did not tie each other, thSince NE and LA did not tie each other, they must have been involved in ties ey must have been involved in ties ey must have been involved in ties ey must have been involved in ties with DC and
with DC and with DC and
with DC and/or/or/or RSL/orRSLRSL. . . . If both NE and LA tied DC, then the 3 NE RSLIf both NE and LA tied DC, then the 3 NE If both NE and LA tied DC, then the 3 NE If both NE and LA tied DC, then the 3 NE matchematchematches would have accounted for 4 of the goals, the matches would have accounted for 4 of the goals, the s would have accounted for 4 of the goals, the s would have accounted for 4 of the goals, the 2 remaining DC
2 remaining DC 2 remaining DC
2 remaining DC matchematchematches would have accounted for 3 of the goals, and the RSL vs LA matches would have accounted for 3 of the goals, and the RSL vs LA s would have accounted for 3 of the goals, and the RSL vs LA s would have accounted for 3 of the goals, and the RSL vs LA matchmatchmatchmatch would involved would involved would involved 2222 goals would involved goals goals goals being scored, which creates a contradiction at 2
being scored, which creates a contradiction at 2 being scored, which creates a contradiction at 2
being scored, which creates a contradiction at 2----0, 10, 10, 10, 1----1, or 01, or 01, or 0----2. If both NE and LA tied RSL, then DC would have to 1, or 02. If both NE and LA tied RSL, then DC would have to 2. If both NE and LA tied RSL, then DC would have to 2. If both NE and LA tied RSL, then DC would have to have had 2 wins and 1 loess, with scores 0
have had 2 wins and 1 loess, with scores 0 have had 2 wins and 1 loess, with scores 0
have had 2 wins and 1 loess, with scores 0----1, 11, 11, 11, 1----0, and 10, and 10, and 1----0; however, this would force a situation of 3 goals being 0, and 10; however, this would force a situation of 3 goals being 0; however, this would force a situation of 3 goals being 0; however, this would force a situation of 3 goals being scored in the
scored in the scored in the
scored in the tie mtie mtie match between RSL and LA, thus creating a contradiction. So, DC and RSL must have each tied one tie match between RSL and LA, thus creating a contradiction. So, DC and RSL must have each tied one atch between RSL and LA, thus creating a contradiction. So, DC and RSL must have each tied one atch between RSL and LA, thus creating a contradiction. So, DC and RSL must have each tied one of NE and LA. If DC had any ties,
of NE and LA. If DC had any ties, of NE and LA. If DC had any ties,
of NE and LA. If DC had any ties, they had two ties and one win. Therefore, they must have tied NE, since they did they had two ties and one win. Therefore, they must have tied NE, since they did they had two ties and one win. Therefore, they must have tied NE, since they did they had two ties and one win. Therefore, they must have tied NE, since they did not beat NE.
not beat NE. not beat NE.
not beat NE. Similarly, if Similarly, if Similarly, if Similarly, if RSL RSL RSL RSL had any ties, thad any ties, thad any ties, thad any ties, they had two hey had two hey had two ties and one loss. Therefore, they must have tied LA, since hey had two ties and one loss. Therefore, they must have tied LA, since ties and one loss. Therefore, they must have tied LA, since ties and one loss. Therefore, they must have tied LA, since they did not lose to LA.
they did not lose to LA. they did not lose to LA.
they did not lose to LA. It follows that DC and RSL must have tied each other. It follows that DC and RSL must have tied each other. It follows that DC and RSL must have tied each other. So, it must be the case that NE tied DC It follows that DC and RSL must have tied each other. So, it must be the case that NE tied DC So, it must be the case that NE tied DC So, it must be the case that NE tied DC and beat RSL and LA, DC tied RSL and beat LA, and RSL tied LA. Th
and beat RSL and LA, DC tied RSL and beat LA, and RSL tied LA. Th and beat RSL and LA, DC tied RSL and beat LA, and RSL tied LA. Th
and beat RSL and LA, DC tied RSL and beat LA, and RSL tied LA. The e e e 3333 NE NE NE NE matchematchematchematches involved a total of 4 goals being s involved a total of 4 goals being s involved a total of 4 goals being s involved a total of 4 goals being scored (since none were scored against NE). The remaining two DC
scored (since none were scored against NE). The remaining two DC scored (since none were scored against NE). The remaining two DC
scored (since none were scored against NE). The remaining two DC matchematchematchematches involved 3 goals being scored (since s involved 3 goals being scored (since s involved 3 goals being scored (since s involved 3 goals being scored (since DC tied NE, 0 to 0). The only remaining
DC tied NE, 0 to 0). The only remaining DC tied NE, 0 to 0). The only remaining
DC tied NE, 0 to 0). The only remaining matchmatchmatch, a tie between RSL and LA, had the only other 2 goamatch, a tie between RSL and LA, had the only other 2 goa, a tie between RSL and LA, had the only other 2 goa, a tie between RSL and LA, had the only other 2 goals, and was thus ls, and was thus ls, and was thus ls, and was thus tied 1 to 1. Since RSL’s lone goal was against LA, and since DC tied RSL,
tied 1 to 1. Since RSL’s lone goal was against LA, and since DC tied RSL, tied 1 to 1. Since RSL’s lone goal was against LA, and since DC tied RSL,
tied 1 to 1. Since RSL’s lone goal was against LA, and since DC tied RSL, the DC vs RSL the DC vs RSL the DC vs RSL the DC vs RSL match match match match ended 0 to 0ended 0 to 0ended 0 to 0.... ended 0 to 0 22228888. . . . (c) (c) (c) (c) Hh
Let M be a miss and H be a hit. The outcomes which would result in Lara winning would be MH, MMMH, Let M be a miss and H be a hit. The outcomes which would result in Lara winning would be MH, MMMH, Let M be a miss and H be a hit. The outcomes which would result in Lara winning would be MH, MMMH, Let M be a miss and H be a hit. The outcomes which would result in Lara winning would be MH, MMMH, MMMMMH
MMMMMH MMMMMH
MMMMMH… in other words, Lara will win if the first hit occurs on an even… in other words, Lara will win if the first hit occurs on an even… in other words, Lara will win if the first hit occurs on an even----numbered shot. P(MH) = … in other words, Lara will win if the first hit occurs on an evennumbered shot. P(MH) = numbered shot. P(MH) = numbered shot. P(MH) = H. ∙∙∙∙ /. ; ; ; ; P(MMMH) =
P(MMMH) = P(MMMH) =
P(MMMH) = H. ∙∙∙∙ H. ∙∙∙∙ H. ∙∙∙∙ /. ; P(MMMMMH) = ; P(MMMMMH) = ; P(MMMMMH) = ; P(MMMMMH) = H. ∙∙∙∙ .H ∙∙∙∙ H. ∙∙∙∙ H. ∙∙∙∙ H. ∙∙∙∙ /. ; this creates a geometric series with first term A = ; this creates a geometric series with first term A = ; this creates a geometric series with first term A = ; this creates a geometric series with first term A = /.1 and common ratio R =
and common ratio R = and common ratio R =
and common ratio R = /.F. The sum of an infinite geometric series, where R is between 0 and 1, is . The sum of an infinite geometric series, where R is between 0 and 1, is . The sum of an infinite geometric series, where R is between 0 and 1, is . The sum of an infinite geometric series, where R is between 0 and 1, is 0 X ƒ_ , or , or , or , or /.1
0 X /.F = = = = H h....
22229999. . . . (e) (e) (e) (e) -F
This is a layered geometric series. There is one series with first term A = This is a layered geometric series. There is one series with first term A = This is a layered geometric series. There is one series with first term A =
This is a layered geometric series. There is one series with first term A = 0- , common ratio R = , common ratio R = , common ratio R = , common ratio R = 0-, added to a series , added to a series , added to a series , added to a series with first term A =
with first term A = with first term A =
with first term A = 010 , common ratio R = , common ratio R = , common ratio R = , common ratio R = 0-, added to a series with first term A = , added to a series with first term A = , added to a series with first term A = , added to a series with first term A = 1-0 , common ratio R = , common ratio R = , common ratio R = , common ratio R = 0-, and so , and so , and so , and so on. Since the sum of an infinite geometric series, where R is between 0 and 1, is
on. Since the sum of an infinite geometric series, where R is between 0 and 1, is on. Since the sum of an infinite geometric series, where R is between 0 and 1, is
on. Since the sum of an infinite geometric series, where R is between 0 and 1, is 0 X ƒ_ , then the sums of these , then the sums of these , then the sums of these , then the sums of these individual series are
individual series are individual series are
individual series are 0H, , , , 0/0, , , , -h0, … which itself represents a geometric series which totals to , … which itself represents a geometric series which totals to , … which itself represents a geometric series which totals to , … which itself represents a geometric series which totals to -F.... 30
30 30 30. . . . (c) 6(c) 6(c) 6 (c) 6
Let f(x) = (x Let f(x) = (x Let f(x) = (x
Let f(x) = (x –––– a)(x a)(x a)(x a)(x –––– b)(x b)(x b)(x b)(x –––– c)(x c)(x c)(x c)(x –––– d), where a, b, c, and d are the four roots of the polynomiald), where a, b, c, and d are the four roots of the polynomiald), where a, b, c, and d are the four roots of the polynomial. Multiplying these d), where a, b, c, and d are the four roots of the polynomial. Multiplying these . Multiplying these . Multiplying these four factors out, the coefficient of the 3
four factors out, the coefficient of the 3 four factors out, the coefficient of the 3
four factors out, the coefficient of the 3rdrdrdrd degree term would be degree term would be degree term would be degree term would be ––––(a + b + c + d). From the function itself, this sum (a + b + c + d). From the function itself, this sum (a + b + c + d). From the function itself, this sum (a + b + c + d). From the function itself, this sum
is equal to is equal to is equal to
is equal to ----6. So, (a + b + c + d) = 6.6. So, (a + b + c + d) = 6.6. So, (a + b + c + d) = 6. 6. So, (a + b + c + d) = 6. 31.
31. 31. 31. (d) (d) (d) (d) 01H
The probability that a crew member is an electrician with a The probability that a crew member is an electrician with a The probability that a crew member is an electrician with a
The probability that a crew member is an electrician with a degree is (0.2)(0.6) = 0.12. By the same reasoning, the degree is (0.2)(0.6) = 0.12. By the same reasoning, the degree is (0.2)(0.6) = 0.12. By the same reasoning, the degree is (0.2)(0.6) = 0.12. By the same reasoning, the probability that a crew member is a plumber with a degree is 0.12, and the probability that a crew member is an probability that a crew member is a plumber with a degree is 0.12, and the probability that a crew member is an probability that a crew member is a plumber with a degree is 0.12, and the probability that a crew member is an probability that a crew member is a plumber with a degree is 0.12, and the probability that a crew member is an engineer with a degree is 0.40. The probability that a crew member has a degree, then,
engineer with a degree is 0.40. The probability that a crew member has a degree, then, engineer with a degree is 0.40. The probability that a crew member has a degree, then,
engineer with a degree is 0.40. The probability that a crew member has a degree, then, is 0.12 + 0.12 + 0.40 = is 0.12 + 0.12 + 0.40 = is 0.12 + 0.12 + 0.40 = is 0.12 + 0.12 + 0.40 = 0.64. The probability that a crew member with a degree is an electrician is
32. (b 32. (b 32. (b 32. (b) 95) 95) 95) 95
First, the foundation of the problem is that the two shortest legs of a triangle must have a sum greater than the First, the foundation of the problem is that the two shortest legs of a triangle must have a sum greater than the First, the foundation of the problem is that the two shortest legs of a triangle must have a sum greater than the First, the foundation of the problem is that the two shortest legs of a triangle must have a sum greater than the long
long long
longest leg. The pole of length 1 cannot be used at all for this reason. Let X be the length of the smallest pole in est leg. The pole of length 1 cannot be used at all for this reason. Let X be the length of the smallest pole in est leg. The pole of length 1 cannot be used at all for this reason. Let X be the length of the smallest pole in est leg. The pole of length 1 cannot be used at all for this reason. Let X be the length of the smallest pole in your triangle, and Y be the difference between the lengths of the middle and longest poles. For example, the your triangle, and Y be the difference between the lengths of the middle and longest poles. For example, the your triangle, and Y be the difference between the lengths of the middle and longest poles. For example, the your triangle, and Y be the difference between the lengths of the middle and longest poles. For example, the triangle made with pole lengths 4,
triangle made with pole lengths 4, triangle made with pole lengths 4,
triangle made with pole lengths 4, 8, 10 would have (X, Y) = (4, 2). X must be no greater than 10. The chart values 8, 10 would have (X, Y) = (4, 2). X must be no greater than 10. The chart values 8, 10 would have (X, Y) = (4, 2). X must be no greater than 10. The chart values 8, 10 would have (X, Y) = (4, 2). X must be no greater than 10. The chart values indicate the number of triangles which can be created for a given (X, Y). The sum of all possible triangles is 95. indicate the number of triangles which can be created for a given (X, Y). The sum of all possible triangles is 95. indicate the number of triangles which can be created for a given (X, Y). The sum of all possible triangles is 95. indicate the number of triangles which can be created for a given (X, Y). The sum of all possible triangles is 95.
X = 2 X = 2 X = 2
X = 2 X = 3X = 3X = 3X = 3 X = 4X = 4X = 4X = 4 X = 5X = 5X = 5X = 5 X = 6 X = 6X = 6X = 6 X = 7X = 7X = 7X = 7 X = 8 X = 8X = 8X = 8 X = 9X = 9 X = 9X = 9 X = 10X = 10X = 10X = 10 YYYY = 1= 1= 1= 1 9999 8888 7777 6666 5555 4444 3333 2222 1111 Y = 2
Y = 2 Y = 2
Y = 2 7777 6666 5555 4444 3333 2222 1111
Y = 3 Y = 3 Y = 3
Y = 3 5555 4444 3333 2222 1111
Y = 4 Y = 4 Y = 4
Y = 4 3333 2222 1111
Y = 5 Y = 5 Y = 5
Y = 5 1111
33333333. . . . (e) 0(e) 0(e) 0 (e) 0 If If If
If /tFu ==== /tu ; cross; cross----multiplying, 9x; cross; crossmultiplying, 9xmultiplying, 9xmultiplying, 9x2222 = 4y= 4y= 4y= 4y2222. In standard form, 9x. In standard form, 9x. In standard form, 9x. In standard form, 9x2222 –––– 4y4y4y4y2222 = 0, or (3x = 0, or (3x = 0, or (3x = 0, or (3x –––– 2y)(3x + 2y) = 0. 2y)(3x + 2y) = 0. 2y)(3x + 2y) = 0. 2y)(3x + 2y) = 0. One of these One of these One of these One of these
factors must be equal to 0. Since factors must be equal to 0. Since factors must be equal to 0. Since
factors must be equal to 0. Since /tFu = = = = /tu = = ((((12x= = 12x12x12x –––– 8y8y8y8y) ) ) ) = 4(3x = 4(3x = 4(3x –––– 2y), none of these factors can be equal to 0. Thus, = 4(3x 2y), none of these factors can be equal to 0. Thus, 2y), none of these factors can be equal to 0. Thus, 2y), none of these factors can be equal to 0. Thus, (3x + 2y) = 0. By factoring,
(3x + 2y) = 0. By factoring, (3x + 2y) = 0. By factoring,
(3x + 2y) = 0. By factoring, ((((9999x + x + x + 6666yyyy)))) = 3(3x + 2y) = 0.x + = 3(3x + 2y) = 0.= 3(3x + 2y) = 0.= 3(3x + 2y) = 0. 33334444. . . . (a) I(a) I(a) I (a) I
If a new data point is higher than the old mean, the mean must rise, even if only a small amount. The median, on If a new data point is higher than the old mean, the mean must rise, even if only a small amount. The median, on If a new data point is higher than the old mean, the mean must rise, even if only a small amount. The median, on If a new data point is higher than the old mean, the mean must rise, even if only a small amount. The median, on the other hand, may or may not rise, depending on the particular values in the middle of the data. The range will the other hand, may or may not rise, depending on the particular values in the middle of the data. The range will the other hand, may or may not rise, depending on the particular values in the middle of the data. The range will the other hand, may or may not rise, depending on the particular values in the middle of the data. The range will not necessarily increase, as
not necessarily increase, as not necessarily increase, as
not necessarily increase, as the new data point may or may not be the new highest data point. The mean and the the new data point may or may not be the new highest data point. The mean and the the new data point may or may not be the new highest data point. The mean and the the new data point may or may not be the new highest data point. The mean and the median can, given the right set of data, increase exactly the same amount. For example, let the original data be {0, median can, given the right set of data, increase exactly the same amount. For example, let the original data be {0, median can, given the right set of data, increase exactly the same amount. For example, let the original data be {0, median can, given the right set of data, increase exactly the same amount. For example, let the original data be {0, 4444, 8, 12, 8, 12, 8, 12, …, 8, 12, …, …, … 64, 68, 72, 64, 68, 72, 7664, 68, 72, 64, 68, 72, 767676}, which has mean 38 and median}, which has mean 38 and median}, which has mean 38 and median}, which has mean 38 and median 38. By adding 80 to the data set, the mean and median 38. By adding 80 to the data set, the mean and median 38. By adding 80 to the data set, the mean and median 38. By adding 80 to the data set, the mean and median shift to 4
shift to 4 shift to 4 shift to 40.0.0.0. 33335555. . . . (a) (a) (a) (a) -7 X FY-7
Cot x = Cot x = Cot x =
Cot x = lDm bmno b ≥ 1. For the given span, this ratio will be greater than or equal to 1 when x is in the ranges of [0, ≥ 1. For the given span, this ratio will be greater than or equal to 1 when x is in the ranges of [0, ≥ 1. For the given span, this ratio will be greater than or equal to 1 when x is in the ranges of [0, ≥ 1. For the given span, this ratio will be greater than or equal to 1 when x is in the ranges of [0, Y-], [π, ], [π, ], [π, ], [π,
.Y -], [2π, ], [2π, ], [2π, ], [2π,
FY
-], and [3π, 10]. The sum of these ranges is ], and [3π, 10]. The sum of these ranges is ], and [3π, 10]. The sum of these ranges is ], and [3π, 10]. The sum of these ranges is Y - + + + +
Y - + + + +
Y
- + (10 + (10 + (10 + (10 ---- 3π) = 10 3π) = 10 3π) = 10 3π) = 10 –––– FY
-. Dividing this total by 10 . Dividing this total by 10 . Dividing this total by 10 . Dividing this total by 10
gives (a) gives (a) gives (a) gives (a) -7 X FY-7 .... 33336666. . . . (d) (d) (d) (d) Yh
Substituting u = x Substituting u = x Substituting u = x
Substituting u = x2222, du = 2xdx, you get , du = 2xdx, you get Cn‰, du = 2xdx, you get , du = 2xdx, you get }Š‹0 /Œ
0 •/ 9 0
}
0 ddddu = u = u = u = 0/(ŽpoX0} –––– ŽpoX00) = = = = 0 / R
\ /S
\ -T = = = =
\ h. . . .
33337777. . . . (b) II only(b) II only(b) II only(b) II only M M M
M5555 = 5M. Since M is positive, M= 5M. Since M is positive, M= 5M. Since M is positive, M= 5M. Since M is positive, M4444 = 5, and is thus, rational. = 5, and is thus, rational. = 5, and is thus, rational. = 5, and is thus, rational. (e-)H = M= M= M= M12121212 is also, necessarily, rational. The others are is also, necessarily, rational. The others are is also, necessarily, rational. The others are is also, necessarily, rational. The others are
not. not. not. not.
33338888. . . . (c) (c) (c) (c) 0H./
By completing both squares, 2 By completing both squares, 2 By completing both squares, 2
By completing both squares, 2Rb SF/T/ + 3+ 3+ 3+ 3(c S H)/= = = = 0H.
/ –––– B B B B . Since both squares are non. Since both squares are non. Since both squares are non. Since both squares are non----negative, B cannot be negative, B cannot be negative, B cannot be negative, B cannot be
any smaller. any smaller. any smaller. any smaller.
33339999. . . . (b) (b) (b) (b) XWHH
The slope of a polar curve at (r, The slope of a polar curve at (r, The slope of a polar curve at (r,
The slope of a polar curve at (r, ϴϴϴϴ) = ) = ) = ) = ••mno y 9•lDm y••lDm y X•mno y. Using that r’ = . Using that r’ = ---- cos . Using that r’ = . Using that r’ = cos ϴϴϴϴ, sin cos cos , sin , sin , sin \1 = = = = 0/ , and cos , and cos , and cos , and cos \1 = = = = WH/, then the , then the , then the , then the aforementioned ratio simplifies to
aforementioned ratio simplifies to aforementioned ratio simplifies to aforementioned ratio simplifies to XWHH ....
40. (d) Green 40. (d) Green 40. (d) Green 40. (d) Green
(I) implies that if a player has a yellow card, then she also has an orange card. (II) implies that if a player has a blue (I) implies that if a player has a yellow card, then she also has an orange card. (II) implies that if a player has a blue (I) implies that if a player has a yellow card, then she also has an orange card. (II) implies that if a player has a blue (I) implies that if a player has a yellow card, then she also has an orange card. (II) implies that if a player has a blue card, then she also has a yellow and a red card. (IV) implies
card, then she also has a yellow and a red card. (IV) implies card, then she also has a yellow and a red card. (IV) implies
card, then she also has a yellow and a red card. (IV) implies that if a player has an orange card, then she also has a that if a player has an orange card, then she also has a that if a player has an orange card, then she also has a that if a player has an orange card, then she also has a blue card. (V) implies that each player must have at least one of orange, yellow, and blue. Since yellow implies blue card. (V) implies that each player must have at least one of orange, yellow, and blue. Since yellow implies blue card. (V) implies that each player must have at least one of orange, yellow, and blue. Since yellow implies blue card. (V) implies that each player must have at least one of orange, yellow, and blue. Since yellow implies orange, orange implies blue, and blue implies yellow, then every player must have yel
orange, orange implies blue, and blue implies yellow, then every player must have yel orange, orange implies blue, and blue implies yellow, then every player must have yel
orange, orange implies blue, and blue implies yellow, then every player must have yellow, orange and blue. Since low, orange and blue. Since low, orange and blue. Since low, orange and blue. Since each player has blue, each player also has red. Thus, all players have red, orange, yellow and blue.
each player has blue, each player also has red. Thus, all players have red, orange, yellow and blue. each player has blue, each player also has red. Thus, all players have red, orange, yellow and blue.
each player has blue, each player also has red. Thus, all players have red, orange, yellow and blue. IIIIf one player is f one player is f one player is f one player is the only player holding all five colors, then th
the only player holding all five colors, then th the only player holding all five colors, then th
the only player holding all five colors, then that playerat playerat player uniquely hold green, as everybody else has all otherat playeruniquely hold green, as everybody else has all otheruniquely hold green, as everybody else has all otheruniquely hold green, as everybody else has all other colors. colors. colors. colors. (III) implies only that each player has yellow and/or green.
